MedVision ad

exponential function (1 Viewer)

Constip8edSkunk

Joga Bonito
Joined
Apr 15, 2003
Messages
2,397
Location
Maroubra
Gender
Male
HSC
2003
I am having a bit of a mind block here and need some help.

I think this is covered in 4U:

given that (1 + 1/k)^k < e < (1 + k)^k+1 for k E Z+ ,
show that (1 + n)^n / e^n < n! < (1 + n)^(1+n) / e^n where n E Z+
 

Mill

Member
Joined
Feb 13, 2003
Messages
256
Gender
Male
HSC
2002
Seems similar to Question 7 of the 4u HSC in 2002 off the top of my head. I haven't looked at in 18 months though, so that's a bit of a guess.
 

Affinity

Active Member
Joined
Jun 9, 2003
Messages
2,062
Location
Oslo
Gender
Undisclosed
HSC
2003
hey you are doing the same course as me

skunk, follow the hints,

e>(1+1)
e>(1+(1/2))^2
...
e>(1+1/n)^n

we can rewrite as

e > (2/1)
e> (3/2)^2
..
e>((n+1)/n)^n

multiply the inequalities together

notice the products

(2/1)(3/2)...((n+1)/n) 'telescopes' and gives (1+n)

(3/2)(4/3)... = (1+n)/2

...

so
e^n > ([(1+n)^n]/n!

similarly

(2/1)^2 > e

(3/2)^3 > e

...

multiply

gives (n+1)^(n+1)/n! >e^2
 
Last edited:

Grey Council

Legend
Joined
Oct 14, 2003
Messages
1,426
Gender
Male
HSC
2004
should this be in the 'extracurricular' maths forum?

and crappo your good Affinity. I think i've said this before somewhere. :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top